2. The Theory of Noetherian RingsAs we have seen in 1.5/8, a ring is called Noetherian if all its ideals are finitelygenera...
56      2. The Theory of Noetherian Ringsfurther. In such a situation we will show in 2.1/8 that the set of prime idealsp1...
2.1 Primary Decomposition of Ideals       57represent all maximal ideals of RS . Identifying each localization (RS )mi wit...
58      2. The Theory of Noetherian Ringsa decomposition                               a = p n1 ∩ . . . ∩ p nr            ...
2.1 Primary Decomposition of Ideals        59     In particular, the powers mn , n > 0, of any maximal ideal m ⊂ R arem-pr...
60     2. The Theory of Noetherian Rings    Now consider elements a, b ∈ R, a = 0, such that ab = 0 and look at thechain o...
2.1 Primary Decomposition of Ideals              61Proof. The case (i) is trivial, whereas (iii) follows directly from the...
62      2. The Theory of Noetherian Rings    (ii) If R is Noetherian, a prime ideal p ⊂ R belongs to Ass(a) if and only if...
2.1 Primary Decomposition of Ideals   63    Conversely, consider an element z ∈ Z(a) and let x ∈ R − a such thatz ∈ (a : x...
64      2. The Theory of Noetherian RingsProof. Let a = r qi be a minimal primary decomposition and set pi = rad(qi ).    ...
2.1 Primary Decomposition of Ideals        65i = r + 1, . . . , r. Then qi RS is pi RS -primary for i = 1, . . . , r and  ...
66       2. The Theory of Noetherian RingsExercises 1. Show for a decomposable ideal a ⊂ R that rad(a) = a implies Ass (a)...
2.2 Artinian Rings and Modules   67    Clearly, an R-module M is Artinian if and only if every non-empty set ofsubmodules ...
68       2. The Theory of Noetherian Ringsalternative method for proving the Noetherian analogue 1.5/10 of Lemma 2.Just as...
2.2 Artinian Rings and Modules       69Proposition 7. Let R be an Artinian ring. Then R contains only finitely manyprime id...
70       2. The Theory of Noetherian Rings    Now consider the case, where R is Artinian. Then R contains only finitelymany...
2.3 The Artin–Rees Lemma     712.3 The Artin–Rees LemmaWe will prove the Artin–Rees Lemma in order to derive Krull’s Inter...
72      2. The Theory of Noetherian RingsProof. Applying the Artin–Rees Lemma to the submodule M = i∈N ai M ofM , we obtai...
2.3 The Artin–Rees Lemma        73     Now the proof of the Artin–Rees Lemma is easy to achieve. Consider thefiltration (Mi...
74      2. The Theory of Noetherian Rings2.4 Krull DimensionIn order to define the dimension of a ring R, we use strictly a...
2.4 Krull Dimension   75r elements, generating an ideal whose radical coincides with the maximal idealm ⊂ R; see Propositi...
76      2. The Theory of Noetherian Rings                               (i )      (i)                              p0 0 ⊂ ...
2.4 Krull Dimension          77Corollary 8. Let R be a Noetherian local ring with maximal ideal m. Thendim R ≤ dimR/m m/m2...
78       2. The Theory of Noetherian RingsDefinition 12. Let R be a Noetherian local ring with maximal ideal m. A setof ele...
2.4 Krull Dimension     79is a strictly ascending chain of prime ideals of length r + 1. From this we seethat dim R X ≥ di...
80      2. The Theory of Noetherian Ringssending Xn onto its residue class in K X1 , . . . , Xn /m. Since on the left-hand...
2.4 Krull Dimension      81Proposition 19. Let R be a regular Noetherian local ring. Then R is an integraldomain.Proof. We...
82       2. The Theory of Noetherian Rings2. Let K X, Y be the polynomial ring in two variables over a field K. Show   dim ...
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Algebraic geometry and commutative algebra

  1. 1. 2. The Theory of Noetherian RingsAs we have seen in 1.5/8, a ring is called Noetherian if all its ideals are finitelygenerated or, equivalently by 1.5/9, if its ideals satisfy the ascending chain con-dition. The aim of the present chapter is to show that the Noetherian hypothesis,as simple as it might look, nevertheless has deep impacts on the structure ofideals and their inclusions, culminating in the theory of Krull dimension, to bedealt with in Section 2.4. To discuss some standard examples of Noetherian and non-Noetherian rings,recall from Hilbert’s Basis Theorem 1.5/14 that all polynomial rings of typeR X1 , . . . , Xn in finitely many variables X1 , . . . , Xn over a Noetherian ring Rare Noetherian. The result extends to algebras of finite type over a Noethe-rian ring R, i.e. R-algebras of type R X1 , . . . , Xn /a where a is an ideal inR X1 , . . . , Xn . In particular, algebras of finite type over a field K or over thering of integers Z are Noetherian. One also knows that all rings of integral alge-braic numbers in finite extensions of Q are Noetherian (use Atiyah–Macdonald[2], 5.17), whereas the integral closure of Z in any infinite algebraic extension ofQ is not; see Section 3.1 for the notion of integral dependence and in particular3.1/8 for the one of integral closure. Also note that any polynomial ring R Xin an infinite family of variables X over a non-zero ring R will not be Noethe-rian. Other interesting examples of non-Noetherian rings belong to the class of(general) valuation rings, as introduced in 9.5/13. To approach the subject of Krull dimension for Noetherian rings, the tech-nique of primary decomposition, developed in Section 2.1, is used as a key tool.We will show in 2.1/6 that such a primary decomposition exists for all ideals aof a Noetherian ring R. It is of type(∗) a = q 1 ∩ . . . ∩ qrwhere q1 , . . . , qr are so-called primary ideals in R. Primary ideals generalizethe notion of prime powers in principal ideal domains, whereas the concept ofprimary decomposition generalizes the one of prime factorization. Looking at a primary decomposition (∗), the nilradicals pi = rad(qi ) are ofparticular significance; they are prime in R and we say that qi is pi -primary.As any finite intersection of p-primary ideals, for any prime ideal p ⊂ R, isp-primary again (see 2.1/4), we may assume that all p1 , . . . , pr belonging tothe primary decomposition (∗) are different. In addition, we can require thatthe decomposition (∗) is minimal in the sense that it cannot be shortened anyS. Bosch, Algebraic Geometry and Commutative Algebra, Universitext, 55DOI 10.1007/978-1-4471-4829-6_2, © Springer-Verlag London 2013
  2. 2. 56 2. The Theory of Noetherian Ringsfurther. In such a situation we will show in 2.1/8 that the set of prime idealsp1 , . . . , pr is uniquely determined by a; it is denoted by Ass(a), referring to themembers of this set as the prime ideals associated to a. There is a uniquenessassertion for some of the primary ideals qi as well (see 2.1/15), although not allof them will be unique in general. Now look at the primary decomposition (∗) and pass to nilradicals, therebyobtaining the decomposition rad(a) = p p∈Ass(a)for the nilradical of a. This is again a primary decomposition, but maybe nota minimal one. Anyway, we know already from 1.3/6 that rad(a) equals theintersection of all prime ideals in R containing a or even better, of all minimalprime divisors of a; the latter are the prime ideals p ⊂ R that are minimal withrespect to the inclusion a ⊂ p. Using 1.3/8, it follows that the minimal primedivisors of a all belong to Ass(a). Thus, their number is finite since Ass(a) isfinite; see 2.1/12. It is this finiteness assertion, which is of utmost importance inthe discussion of Krull dimensions for Noetherian rings. Translated to the worldof schemes it corresponds to the fact that every Noetherian scheme consists ofonly finitely many irreducible components; see 7.5/5. To give an application of the just explained finiteness of sets of associatedprime ideals, consider a Noetherian ring R where every prime ideal is maximal;we will say that R is of Krull dimension 0. Then all prime ideals of R areassociated to the zero ideal in R and, hence, there can exist only finitely manyof them. Using this fact in conjunction with some standard arguments, we canshow in 2.2/8 that R satisfies the descending chain condition for ideals and,thus, is Artinian. Conversely, it is shown that every Artinian ring is Noetherianof Krull dimension 0. The Krull dimension of a general ring R is defined as the supremum of alllengths n of chains of prime ideals p0 . . . pn in R and is denoted by dim R.Restricting to chains ending at a given prime ideal p ⊂ R, the correspondingsupremum is called the height of p, denoted by ht p. As a first major resultin dimension theory we prove Krull’s Dimension Theorem 2.4/6, implying thatht p is finite if R is Noetherian. From this we conclude that the Krull dimensionof any Noetherian local ring is finite; see 2.4/8. On the other hand, it is not toohard to construct (non-local) Noetherian rings R where dim R = ∞. Namely,following Nagata [22], Appendix A1, Example 1, we consider a polynomial ringR = K X1 , X2 , . . . over a field K where each Xi is a finite system of variables,say of length ni , such that limi ni = ∞. Let pi be the prime ideal that isgenerated by Xi in R and let S ⊂ R be the multiplicative system given by thecomplement of the union ∞ pi . Then we claim that the localization RS is a i=1Noetherian ring of infinite dimension. Indeed, looking at residue rings of typeR/(Xr , Xr+1 , . . .) for sufficiently large indices r, we can use 1.3/7 in order toshow that the pi are just those ideals in R that are maximal with respect to theproperty of being disjoint from S. Therefore the ideals mi = pi RS , i = 1, 2, . . .,
  3. 3. 2.1 Primary Decomposition of Ideals 57represent all maximal ideals of RS . Identifying each localization (RS )mi withthe localization K(Xj ; j = i) Xi (Xi ) , we see with the help of Hilbert’s BasisTheorem 1.5/14 that all these rings are Noetherian. It follows, as remarkedafter defining Krull dimensions in 2.4/2, that dim(RS )mi = ht(mi ) ≥ ni . Hence,we get dim RS = ∞. To show that RS is Noetherian, indeed, one can use thefacts that all localizations (RS )mi are Noetherian and that any non-zero elementa ∈ RS is contained in at most finitely many of the maximal ideals mi ⊂ RS .See the reference of Nagata [22] given above in conjunction with Exercise 4.3/5. Krull’s Dimension Theorem 2.4/6 reveals a very basic fact: if R is a Noethe-rian ring and a ⊂ R an ideal generated by r elements, then ht p ≤ r for everyminimal prime divisor p of a. For the proof we need the finiteness of Ass(a) asdiscussed in 2.1/12 and, within the context of localizations, the characterizationof Noetherian rings of dimension 0 in terms of Artinian rings. Another techni-cal ingredient is Krull’s Intersection Theorem 2.3/2, which in turn is based onNakayama’s Lemma 1.4/10 and the Artin–Rees Lemma 2.3/1. Both, Krull’sIntersection Theorem and the Artin–Rees Lemma allow nice topological inter-pretations in terms of ideal-adic topologies; see the corresponding discussion inSection 2.3. For Noetherian local rings there is a certain converse of Krull’s IntersectionTheorem. Consider such a ring R of dimension d, and let m be its maximalideal so that ht m = d. Then every m-primary ideal q ⊂ R satisfies rad(q) = mand, thus, by Krull’s Dimension Theorem, cannot be generated by less thand elements. On the other hand, a simple argument shows in 2.4/11 that therealways exist m-primary ideals in R that are generated by a system of d elements.Alluding to the situation of polynomial rings over fields, such systems are calledsystems of parameters of the local ring R. Using parameters, the dimensiontheory of Noetherian local rings can be handled quite nicely; see for examplethe results 2.4/13 and 2.4/14. Furthermore, we can show that a polynomial ringR X1 , . . . , Xn in finitely many variables X1 , . . . , Xn over a Noetherian ring Rhas dimension dim R + n, which by examples of Seidenberg may fail to be trueif R is not Noetherian any more. A very particular class of Noetherian local rings is given by the subclass ofregular local rings, where a Noetherian local ring is called regular if its maxi-mal ideal can be generated by a system of parameters. Such rings are integraldomains, as we show in 2.4/19. They are quite close to (localizations of) poly-nomial rings over fields and are useful to characterize the geometric notion ofsmoothness in Algebraic Geometry; see for example 8.5/15.2.1 Primary Decomposition of IdealsLet R be a principal ideal domain. Then R is factorial and any non-zero elementa ∈ R admits a factorization a = εpn1 . . . pnr with a unit ε ∈ R∗ , pairwise non- 1 requivalent prime elements pi ∈ R, and exponents ni > 0 where these quantitiesare essentially unique. Passing to ideals, it follows that every ideal a ⊂ R admits
  4. 4. 58 2. The Theory of Noetherian Ringsa decomposition a = p n1 ∩ . . . ∩ p nr 1 rwith pairwise different prime ideals pi that are unique up to order, and expo-nents ni > 0 that are unique as well. The purpose of the present section is tostudy similar decompositions for more general rings R where the role of theabove prime powers pni is taken over by the so-called primary ideals. In the ifollowing we start with a general ring R (commutative and with a unit element,as always). Only later, when we want to show the existence of primary decom-positions, R will be assumed to be Noetherian. For a generalization of primarydecompositions to the context of modules see Serre [24], I.B.Definition 1. A proper ideal q ⊂ R is called a primary ideal if ab ∈ q for anyelements a, b ∈ R implies a ∈ q or, if the latter is not the case, that there is anexponent n ∈ N such that bn ∈ q. Clearly, any prime ideal is primary. Likewise, for a prime element p of afactorial ring, all powers (p)n , n > 0, are primary. But for general rings we willsee that the higher powers of prime ideals may fail to be primary. Also notethat an ideal q ⊂ R is primary if and only if the zero ideal in R/q is primary.The latter amounts to the fact that all zero-divisors in R/q are nilpotent. Moregenerally, if π : R E R/a is the canonical projection from R onto its quotientby any ideal a ⊂ R, then an ideal q ⊂ R/a is primary if and only if its preimageπ −1 (q) is primary in R.Remark 2. Let q ⊂ R be a primary ideal. Then p = rad(q) is a prime ideal inR. It is the unique smallest prime ideal p containing q.Proof. First, rad(q) is a proper ideal in R, since the same holds for q. Nowlet ab ∈ rad(q) for some elements a, b ∈ R where a ∈ rad(q). Then there is anexponent n ∈ N such that an bn ∈ q. Since a ∈ rad(q) implies an ∈ q, there existsan exponent n ∈ N such that bnn ∈ q. However, the latter shows b ∈ rad(q) andwe see that rad(q) is prime. Furthermore, if p is a prime ideal in R containingq, it must contain p = rad(q) as well. Hence, the latter is the unique smallestprime ideal containing q. Given a primary ideal q ⊂ R, we will say more specifically that q isp-primary for the prime ideal p = rad(q).Remark 3. Let q ⊂ R be an ideal such that its radical rad(q) coincides with amaximal ideal m ⊂ R. Then q is m-primary.Proof. Replacing R by R/q and m by m/q, we may assume q = 0. Then m is theonly prime ideal in R by 1.3/4 and we see that R is a local ring with maximalideal m. Since R − m consists of units, all zero divisors of R are contained in mand therefore nilpotent. This shows that q = 0 is a primary ideal in R.
  5. 5. 2.1 Primary Decomposition of Ideals 59 In particular, the powers mn , n > 0, of any maximal ideal m ⊂ R arem-primary. However, there might exist m-primary ideals of more general type,as can be read from the example of the polynomial ring R = K X, Y over afield K. Namely, q = (X, Y 2 ) is m-primary for m = (X, Y ), although it is not apower of m. On the other hand, there are rings R admitting a (non-maximal)prime ideal p such that the higher powers of p are not primary. To construct sucha ring, look at the polynomial ring R = K X, Y, Z in three variables over a fieldK. The ideal q = (X 2 , XZ, Z 2 , XY − Z 2 ) ⊂ R has radical p = rad(q) = (X, Z),which is prime. However, q is not primary, since XY ∈ q, although X ∈ qand Y ∈ rad(q) = p. Now observe that that q = (X, Z)2 + (XY − Z 2 ). Since(XY − Z 2 ) ⊂ q ⊂ p, we may pass to the residue ring R = R/(XY − Z 2 ),obtaining q = p2 for the ideals induced from p and q. Then p is prime in R, butp2 cannot be primary since its preimage q ⊂ R is not primary.Lemma 4. Let q1 , . . . , qr be p-primary ideals in R for some prime ideal p ⊂ R.Then the intersection q = r qi is p-primary. i=1Proof. First observe that rad(q) = r rad(qi ) = p, since we have rad(qi ) = p i=1for all i. Next consider elements a, b ∈ R such that ab ∈ q or, in other words,ab ∈ qi for all i. Now if b ∈ rad(q) = p, we get b ∈ rad(qi ) for all i. But then, asall qi are primary, we must have a ∈ qi for all i and, hence, a ∈ q. This showsthat q is p-primary.Definition 5. A primary decomposition of an ideal a ⊂ R is a decomposition r a= qi i=1into primary ideals qi ⊂ R. The decomposition is called minimal if the corre-sponding prime ideals pi = rad(qi ) are pairwise different and the decompositionis unshortenable; the latter means that i=j qi ⊂ qj for all j = 1, . . . , r. An ideal a ⊂ R is called decomposable if it admits a primary decomposition. In particular, it follows from Lemma 4 that any primary decomposition ofan ideal a ⊂ R can be reduced to a minimal one. On the other hand, to show theexistence of primary decompositions, special assumptions are necessary such asR being Noetherian.Theorem 6. Let R be a Noetherian ring. Then any ideal a ⊂ R is decomposable,i.e. it admits a primary decomposition and, in particular, a minimal one.Proof. Let us call a proper ideal q ⊂ R irreducible if from any relation q = a1 ∩a2with ideals a1 , a2 ⊂ R we get q = a1 or q = a2 . In a first step we show thatevery irreducible ideal q ⊂ R is primary. To achieve this we may pass to thequotient R/q, thereby assuming that q is the zero ideal in R.
  6. 6. 60 2. The Theory of Noetherian Rings Now consider elements a, b ∈ R, a = 0, such that ab = 0 and look at thechain of ideals Ann(b) ⊂ Ann(b2 ) ⊂ . . . ⊂ Rwhere the annihilator Ann(bi ) consists of all elements x ∈ R such that xbi = 0.The chain becomes stationary, since R is Noetherian. Hence, there is an indexn ∈ N such that Ann(bi ) = Ann(bn ) for all i ≥ n. We claim that (a) ∩ (bn ) = 0.Indeed, for any y ∈ (a)∩(bn ) there are elements c, d ∈ R such that y = ca = dbn .Then yb = 0 and, hence, dbn+1 = 0 so that d ∈ Ann(bn+1 ) = Ann(bn ). It followsy = dbn = 0 and therefore (a) ∩ (bn ) = 0, as claimed. Assuming that thezero ideal of R is irreducible, the latter relation shows (bn ) = 0, since a = 0.Consequently, b is nilpotent and we see that 0 ⊂ R is primary. It remains to show that every ideal a ⊂ R is a finite intersection of ir-reducible ideals. To do this, let M be the set of all ideals in R that are notrepresentable as a finite intersection of irreducible ideals. If M is non-empty,we can use the fact that R is Noetherian and thereby show that M contains amaximal element a. Then a cannot be irreducible, due to the definition of M .Therefore we can write a = a1 ∩ a2 with ideals a1 , a2 ⊂ R that do not belong toM . But then a1 and a2 are finite intersections of irreducible ideals in R and thesame is true for a. However, this contradicts the definition of a and thereforeyields M = ∅. Thus, every ideal in R is a finite intersection of irreducible and,hence, primary ideals. To give an example of a non-trivial primary decomposition, consider againthe polynomial ring R = K X, Y, Z in three variables over a field K and itsideals p = (X, Y ), p = (X, Z), m = (X, Y, Z).Then a = p · p satisfies a = p ∩ p ∩ m2and, as is easily checked, this is a minimal primary decomposition. In the following we will discuss uniqueness assertions for primary decom-positions of decomposable ideals. In particular, this applies to the Noetheriancase, although the uniqueness results themselves do not require the Noetherianassumption. We will need the notion of ideal quotients, as introduced in Sec-tion 1.1. Recall that for an ideal q ⊂ R and an element x ∈ R the ideal quotient(q : x) consists of all elements a ∈ R such that ax ∈ q.Lemma 7. Consider a primary ideal q ⊂ R as well as the corresponding primeideal p = rad(q). Then, for any x ∈ R, (i) (q : x) = R if x ∈ q, (ii) (q : x) is p-primary if x ∈ q, (iii) (q : x) = q if x ∈ p. Furthermore, if R is Noetherian, there exists an element x ∈ R such that(q : x) = p.
  7. 7. 2.1 Primary Decomposition of Ideals 61Proof. The case (i) is trivial, whereas (iii) follows directly from the definition ofprimary ideals, since p = rad(q). To show (ii), assume x ∈ q and observe that q ⊂ (q : x) ⊂ p = rad(q),where the first inclusion is trivial and the second follows from the fact that qis p-primary. In particular, we can conclude that rad (q : x) = p. Now leta, b ∈ R such that ab ∈ (q : x), and assume b ∈ p. Then abx ∈ q implies ax ∈ qas b ∈ p and, hence, a ∈ (q : x). In particular, (q : x) is p-primary. Next assume that R is Noetherian. Then p = rad(q) is finitely generatedand there exists an integer n ∈ N such that pn ⊂ q. Assume that n is minimalso that pn−1 ⊂ q, and choose x ∈ pn−1 − q. Then p ⊂ (q : x), since p · x ∈ pn ⊂ q,and we see from (ii) that (q : x) ⊂ p. Hence, (q : x) = p.Theorem 8. For a decomposable ideal a ⊂ R, let r a= qi i=1be a minimal primary decomposition. Then the set of corresponding prime idealspi = rad(qi ) is independent of the chosen decomposition and, thus, depends onlyon the ideal a. More precisely, it coincides with the set of all prime ideals in Rthat are of type rad (a : x) for x varying over R.Proof. Let x ∈ R. The primary decomposition of a yields (a : x) = r i=1 (qi : x)and therefore by Lemma 7 a primary decomposition r rad (a : x) = rad (qi : x) = pi . i=1 x∈qiNow assume that the ideal rad (a : x) is prime. Then we see from 1.3/8 that itmust coincide with one of the prime ideals pi . Conversely, fix any of the idealspi , say pj where j ∈ {1, . . . , r}, and choose x ∈ ( i=j qi ) − qj . Then, as we havejust seen, we get the primary decomposition rad (a : x) = pi = pj x∈qiand we are done. The prime ideals p1 , . . . , pr in the situation of Theorem 8 are uniquely as-sociated to the ideal a. We use this fact to give them a special name.Definition and Proposition 9. Let a ⊂ R be an ideal. (i) The set of all prime ideals in R that are of type rad (a : x) for somex ∈ R is denoted by Ass(a); its members are called the prime ideals associatedto a.1 1 Let us point out that usually the set Ass(a) is defined by the condition in (ii), whichis equivalent to (i) only in the Noetherian case. However, we prefer to base the definition of
  8. 8. 62 2. The Theory of Noetherian Rings (ii) If R is Noetherian, a prime ideal p ⊂ R belongs to Ass(a) if and only ifit is of type (a : x) for some x ∈ R. (iii) Ass(a) is finite if R is Noetherian or, more generally, if a is decompos-able.Proof. Assume that R is Noetherian and let a = r qi be a minimal pri- i=1mary decomposition of a; see Theorem 6. Choosing x ∈ ( i=j qi ) − qj for anyj ∈ {1, . . . , r}, we obtain (a : x) = (qi : x) = (qj : x) x∈qiwhere (qj : x) is pj -primary by Lemma 7. Furthermore, by the same result, wecan find an element x ∈ R such that (qj : xx ) = (qj : x) : x = pj . But then,since xx ∈ qi for all i = j, we conclude that (a : xx ) = (qi : xx ) = (qj : xx ) = pj . xx ∈qiThis shows that any prime ideal in Ass(a) is as stated in (ii). Since prime idealsof type (ii) are also of type (i), assertion (ii) is clear. Furthermore, (iii) followsfrom Theorem 8. There is a close relationship between the prime ideals associated to an ideala ⊂ R and the set of zero divisors in R/a. For an arbitrary ideal a ⊂ R let uscall Z(a) = (a : x) = {z ∈ R ; zx ∈ a for some x ∈ R − a}, x∈R−athe set of zero divisors modulo a in R. Indeed, an element z ∈ R belongs toZ(a) if and only if its residue class z ∈ R/a is a zero divisor. Furthermore, if apower z n of some element z ∈ R belongs to Z(a), then z itself must belong toZ(a) and therefore Z(a) = rad (a : x) . x∈R−aCorollary 10. If a is a decomposable ideal in R, then Z(a) = p. p∈Ass(a)Proof. Let p ∈ Ass(a). Then, by the definition of Ass(a), there exists some x ∈ Rsuch that p = rad (a : x) . Since p is a proper ideal in R, the element x cannotbelong to a and, hence, p ⊂ Z(a).Ass(a) on the condition in (i), since this is more appropriate for handling primary decompo-sitions in non-Noetherian situations.
  9. 9. 2.1 Primary Decomposition of Ideals 63 Conversely, consider an element z ∈ Z(a) and let x ∈ R − a such thatz ∈ (a : x). Furthermore, let a = r qi be a minimal primary decomposition i=1of a. Then, writing pi = rad(qi ), we get z ∈ (a : x) = (qi : x) ⊂ pi , x∈qi x∈qisince (qi : x) is pi -primary for x ∈ qi and coincides with R for x ∈ qi ; seeLemma 7. Using x ∈ a = r qi , the set of all i such that x ∈ qi is non-empty i=1and we are done. Although for a minimal primary decomposition a = r qi of some ideal i=1a ⊂ R, proper inclusions of type qi qj are not allowed, these can neverthelessoccur on the level of prime ideals associated to a. For example, look at thepolynomial ring R = K X, Y in two variables over a field K. Then the ideala = (X 2 , XY ) admits the primary decomposition a = p1 ∩ p2 where p1 = (X) 2and p2 = (X, Y ). Clearly, this primary decomposition is minimal and, hence,we get Ass(a) = {p1 , p2 } with p1 p2 .Definition 11. Given any ideal a ⊂ R, the subset of all prime ideals thatare minimal in Ass(a) is denoted by Ass (a) and its members are called theisolated prime ideals associated to a. All other elements of Ass(a) are said tobe embedded prime ideals. The notion of isolated and embedded prime ideals is inspired from geometry.Indeed, passing to the spectrum Spec R of all prime ideals in R and looking atzero sets of type V (E) for subsets E ⊂ R as done in Section 6.1, a strict inclusionof prime ideals p p is reflected on the level of zero sets as a strict inclusionV (p) V (p ). In particular, the zero set V (p ) is “embedded” in the bigger oneV (p) and, likewise, p is said to be an embedded prime ideal associated to aif we have p, p ∈ Ass(a) with p p . Also recall from Theorem 8 that if a isdecomposable with primary decomposition a = r qi , then Ass(a) consists of i=1the finitely many radicals pi = rad(qi ). Since V (qi ) = V (pi ), one obtains V (a) = V (p) = V (p), p∈Ass(a) p∈Ass (a)as explained at the end of Section 1.3.Proposition 12. Let a ⊂ R be an ideal and assume that it is decomposable;for example, the latter is the case by Theorem 6 if R is Noetherian. Then: (i) The set Ass(a) and its subset Ass (a) of isolated prime ideals a are finite. (ii) Every prime ideal p ⊂ R satisfying a ⊂ p contains a member of Ass (a).Consequently, Ass (a) consists of all prime ideals p ⊂ R that are minimal amongthe ones satisfying a ⊂ p ⊂ R. (iii) rad(a) = p∈Ass (a) p is the primary decomposition of the nilradical of a.
  10. 10. 64 2. The Theory of Noetherian RingsProof. Let a = r qi be a minimal primary decomposition and set pi = rad(qi ). i=1Then Ass(a) is finite, since it consists of the prime ideals p1 , . . . , pr by Theo-rem 8. Furthermore, we obtain a ⊂ p for any p ∈ Ass(a). Conversely, considera prime ideal p ⊂ R such that a ⊂ p. Then we see from a = r qi ⊂ p by i=11.3/8 that there is an index i satisfying qi ⊂ p and, hence, pi = rad(qi ) ⊂ p.In particular, p contains an element of Ass(a) and, thus, of Ass (a). This settles(ii), whereas (iii) is derived from the primary decomposition of a by passing toradicals. Looking at a primary decomposition a = r qi of some ideal a ⊂ R, we i=1know from Theorem 8 that the associated prime ideals pi = rad(qi ) are unique;this led us to the definition of the set Ass(a). In the remainder of this sectionwe want to examine the uniqueness of the primary ideals qi themselves. Notall of them will be unique, as is demonstrated by the two different primarydecompositions (X 2 , XY ) = (X) ∩ (X, Y )2 = (X) ∩ (X 2 , Y )in the polynomial ring R = K X, Y over a field K. We start with the followinggeneralization of 1.2/5:Lemma 13. Let S ⊂ R be a multiplicative system and RS the localization of Rby S. Furthermore, consider a prime ideal p ⊂ R and a p-primary ideal q ⊂ R. (i) If S ∩ p = ∅, then qRS = RS . (ii) If S ∩ p = ∅, then pRS is a prime ideal in RS satisfying pRS ∩ R = pand qRS is a pRS -primary ideal in RS satisfying qRS ∩ R = q.Proof. Since p = rad(q), we see that S ∩ p = ∅ implies S ∩ q = ∅ and, hence,qRS = RS by 1.2/5. This settles (i). To show (ii), assume that S ∩ p = ∅. Thenwe conclude from 1.2/5 that pRS is a prime ideal in RS containing qRS andsatisfying pRS ∩ R = p. In particular, qRS is a proper ideal in RS and we caneven see that pRS = rad(qRS ). To show qRS ∩ R = q, observe first that q ⊂ qRS ∩ R for trivial reasons.Conversely, consider an element a ∈ qRS ∩ R. Then there are elements a ∈ qand s ∈ S such that a = a in RS . So there exists t ∈ S such that (as − a )t = 0 1 sin R and, hence, ast ∈ q. Since st ∈ S, no power of it will be contained in q andwe get a ∈ q from the primary ideal condition of q. Therefore qRS ∩ R = q, asclaimed. It remains to show that qRS is a pRS -primary ideal. Indeed, look atelements a , a ∈ RS such that aa ∈ qRS , and assume a ∈ qRS . Then aa ∈ qRS s s ss s 1and, hence, aa ∈ q, as we just have shown. Furthermore, a ∈ qRS implies a ∈ q sso that a ∈ p. But then a ∈ pRS and we are done. sProposition 14. Let S ⊂ R be a multiplicative system and RS the localizationof R by S. Furthermore, consider a minimal primary decomposition a = r qi i=1of some ideal a ⊂ R. Let pi = rad(qi ) for i = 1, . . . , r and assume that r is aninteger, 0 ≤ r ≤ r, such that pi ∩ S = ∅ for i = 1, . . . , r and pi ∩ S = ∅ for
  11. 11. 2.1 Primary Decomposition of Ideals 65i = r + 1, . . . , r. Then qi RS is pi RS -primary for i = 1, . . . , r and r r (i) aRS = qi RS as well as (ii) aRS ∩ R = qi i=1 i=1are minimal primary decompositions in RS and R.Proof. The equations (i) and (ii) can be deduced from Lemma 13 if we knowthat the formation of extended ideals in RS commutes with finite intersections.However, to verify this is an easy exercise. If b, c are ideals in R, then clearly(b ∩ c)RS ⊂ bRS ∩ cRS . To show the reverse inclusion, consider an elementbs = s ∈ bRS ∩ cRS where b ∈ b, c ∈ c, and s, s ∈ S. Then there is some ct ∈ S such that (bs − cs)t = 0 and, hence, bs t = cst ∈ b ∩ c. But thenbs = ss t ∈ (b ∩ c)RS , which justifies our claim. bs t Now we see from Lemma 13 that (i) and (ii) are primary decompositions.The primary decomposition (ii) is minimal, since the decomposition of a westarted with was assumed to be minimal. But then it is clear that the decompo-sition (i) is minimal as well. Indeed, the pi RS for i = 1, . . . , r are prime idealssatisfying pi RS ∩ R = pi . So they must be pairwise different, since p1 , . . . , pr arepairwise different. Furthermore, the decomposition (i) is unshortenable, sincethe same is true for (ii) and since qi RS ∩ R = qi for i = 1, . . . , r by Lemma 13. Now we are able to derive the following strengthening of the uniquenessassertion in Theorem 8:Theorem 15. Let a ⊂ R be a decomposable ideal and a = r qi a minimal i=1primary decomposition where qi is pi -primary. Then qi = aRpi ∩R for all indicesi such that pi ∈ Ass (a). In particular, a member qi of the primary decomposition of a is unique,depending only on a and the prime ideal pi = rad(qi ) ∈ Ass(a), if pi belongs toAss (a). In the latter case, we call qi the pi -primary part of a.Proof. Consider an index j ∈ {1, . . . , r} such that pj = rad(qj ) ∈ Ass (a) andset S = R − pj . Then S ∩ pj = ∅, while S ∩ pi = ∅ for all i = j. Therefore wecan conclude from Proposition 14 that aRS ∩ R = qj . Let R be a Noetherian ring. As an application to the above uniquenessresult we can define for n ≥ 1 the nth symbolic power of any prime ideal p ⊂ R.Namely, consider a primary decomposition of pn , which exists by Theorem 6.Then Ass (a) = {p} by Proposition 12 and the p-primary part of pn is unique,given by pn Rp ∩ R. The latter is denoted by p(n) and called the nth symbolicpower of p. Of course, pn ⊂ p(n) and we have pn = p(n) if and only if pn isprimary itself.
  12. 12. 66 2. The Theory of Noetherian RingsExercises 1. Show for a decomposable ideal a ⊂ R that rad(a) = a implies Ass (a) = Ass(a). Is the converse of this true as well? 2. Let p ⊂ R be a prime ideal and assume that R is Noetherian. Show that the nth symbolic power p(n) is the smallest p-primary ideal containing pn . Can we expect that there is a smallest primary ideal containing pn ? 3. For a non-zero Noetherian ring R let p1 , . . . , pr be its (pairwise different) prime ideals that are associated to the zero ideal 0 ⊂ R. Show that there is an integer (n) n0 ∈ N such that r pi = 0 for all n ≥ n0 and, hence, that the latter are i=1 (n) (n ) minimal primary decompositions of the zero ideal. Conclude that pi = pi 0 for (n) all n ≥ n0 if pi is an isolated prime ideal and show that the symbolic powers pi are pairwise different if pi is embedded. 4. Consider a decomposable ideal a ⊂ R and a multiplicative subset S ⊂ R such that S ∩ a = ∅. Then the map p E p ∩ R from prime ideals p ⊂ RS to prime ideals in R restricts to an injective map Ass(aRS ) ⊂ E Ass(a). 5. Let R = K X1 , . . . Xr be the polynomial ring in finitely many variables over a field K and let p ⊂ R be an ideal that is generated by some of the variables Xi . Show that p is prime and that all its powers pn , n ≥ 1, are p-primary. In particular, p(n) = pn for n ≥ 1. 6. Let M be an R-module. A submodule N ⊂ M is called primary if N = M and if the multiplication by any element a ∈ R is either injective or nilpotent on M/N ; in other words, if there exists an element x ∈ M − N such that ax ∈ N , then there is an exponent n ≥ 1 satisfying an M ⊂ N . (a) Let (N : M ) = {a ∈ R ; aM ⊂ N } and show that p = rad(N : M ) is a prime ideal in R if N is primary in M ; we say that N is p-primary in M . (b) Show that an ideal q ⊂ R, viewed as a submodule of R, is p-primary in the sense of (a) for some prime ideal p ⊂ R if and only if it is p-primary in the sense of Definition 1. (c) Generalize the theory of primary decompositions from ideals to submodules of M ; see for example Serre [24], I.B.2.2 Artinian Rings and ModulesAs we have seen in 1.5/9, a module M over a ring R is Noetherian if and onlyif it satisfies the ascending chain condition, i.e. if and only if every ascendingchain of submodules M1 ⊂ M2 ⊂ . . . ⊂ M becomes stationary. Switching todescending chains we arrive at the notion of Artinian modules.Definition 1. An R-module M is called Artinian if every descending chain ofsubmodules M ⊃ M1 ⊃ M2 ⊃ . . . becomes stationary, i.e. if there is an index i0such that Mi0 = Mi for all i ≥ i0 . A ring R is called Artinian if, viewed as anR-module, it is Artinian.
  13. 13. 2.2 Artinian Rings and Modules 67 Clearly, an R-module M is Artinian if and only if every non-empty set ofsubmodules admits a minimal element. In a similar way, Noetherian modulesare characterized by the existence of maximal elements. This suggests a closeanalogy between Noetherian and Artinian modules. However, let us point outright away that such an analogy does not exist, as is clearly visible on the levelof rings. Namely, the class of Artinian rings is, in fact, a subclass of the oneof Noetherian rings, as we will prove below in Theorem 8. Thus, for rings theascending chain condition is, in fact, a consequence of the descending chaincondition. Looking at some simple examples, observe that any field is Artinian. Fur-thermore, a vector space over a field is Artinian if and only if it is of finitedimension. Since, for any prime p, the ring of integers Z admits the strictlydescending sequence of ideals Z (p) (p2 ) . . ., we see that Z cannotbe Artinian. However, any quotient Z/nZ for some n = 0 is Artinian, since itconsists of only finitely many elements. For the discussion of general properties of Artinian modules, the Artiniananalogue of 1.5/10 is basic: f gLemma 2. Let 0 E M E M E M E 0 be an exact sequence ofR-modules. Then the following conditions are equivalent: (i) M is Artinian. (ii) M and M are Artinian.Proof. If M is Artinian, M is Artinian as well, since any chain of submodules inM can be viewed as a chain of submodules in M . Furthermore, given a chain ofsubmodules in M , its preimage in M becomes stationary if M is Artinian. Butthen the chain we started with in M must become stationary as well. Thus, itis clear that condition (i) implies (ii). Conversely, assume that M and M are Artinian. Then, for any descendingchain of submodules M ⊃ M1 ⊃ M2 ⊃ . . ., we know that the chains M ⊃ f −1 (M1 ) ⊃ f −1 (M2 ) ⊃ . . . , M ⊃ g(M1 ) ⊃ g(M2 ) ⊃ . . .become stationary, say at some index i0 , and we can look at the commutativediagrams 0 E f −1 (Mi ) E Mi E g(Mi ) E0 ∩ σi c 0 E f −1 (Mi0 ) E Mi0 E g(Mi0 ) E 0for i ≥ i0 . Since the rows are exact, all inclusions σi : Mi E Mi must be 0bijective. Hence, the chain M ⊃ M1 ⊃ M2 ⊃ . . . becomes stationary at i0 aswell and we see that M is Artinian. Using the fact that Noetherian modules can be characterized by the ascend-ing chain condition, as shown in 1.5/9, the above line of arguments yields an
  14. 14. 68 2. The Theory of Noetherian Ringsalternative method for proving the Noetherian analogue 1.5/10 of Lemma 2.Just as in the Noetherian case, let us derive some standard consequences fromthe above lemma.Corollary 3. Let M1 , M2 be Artinian R-modules. Then: (i) M1 ⊕ M2 is Artinian. (ii) If M1 , M2 are given as submodules of some R-module M , then M1 + M2and M1 ∩ M2 are Artinian.Proof. Due to Lemma 2, the proof of 1.5/11 carries over literally, just replacingNoetherian by Artinian.Corollary 4. For a ring R, the following conditions are equivalent: (i) R is Artinian. (ii) Every R-module M of finite type is Artinian.Proof. Use the argument given in the proof of 1.5/12. There is a straightforward generalization of Lemma 2 to chains of modules.Proposition 5. Let 0 = M0 ⊂ M1 ⊂ . . . ⊂ Mn = M be a chain of submodulesof an R-module M . Then M is Artinian if and only if all quotients Mi /Mi−1 ,i = 1, . . . , n, are Artinian. The same is true for Artinian replaced by Noetherian.Proof. The only-if part is trivial. To prove the if part we use induction by n,observing that the case n = 1 is trivial. Therefore let n > 1 and assume byinduction hypothesis that Mn−1 is Artinian (resp. Noetherian). Then the exactsequence 0 E Mn−1 EM E M/Mn−1 E0shows that M is Artinian by Lemma 2 (resp. Noetherian by 1.5/10).Corollary 6. Let R be a ring with (not necessarily distinct) maximal idealsm1 , . . . , mn ⊂ R such that m1 · . . . · mn = 0. Then R is Artinian if and only if itis Noetherian.Proof. Look at the chain of ideals 0 = m1 · . . . · mn ⊂ . . . ⊂ m1 · m2 ⊂ m1 ⊂ Rand, for i = 1, . . . , n, consider the quotient Vi = (m1 · . . . · mi−1 )/(m1 · . . . · mi )as an R/mi -vector space. Being a vector space, Vi is Artinian if and only if it isof finite dimension and the latter is equivalent to Vi being Noetherian. As thereis no essential difference between Vi as an R/mi -module or as an R-module,Proposition 5 shows that R is Artinian if and only if it is Noetherian.
  15. 15. 2.2 Artinian Rings and Modules 69Proposition 7. Let R be an Artinian ring. Then R contains only finitely manyprime ideals. All of them are maximal and, hence, minimal as well. In particular,the Jacobson radical j(R) coincides with the nilradical rad(R) of R.Proof. Let p ⊂ R be a prime ideal. In order to show that p is a maximal ideal, wehave to show that the quotient R/p is a field. So consider a non-zero elementx ∈ R/p. Since R/p is Artinian, the chain (x1 ) ⊃ (x2 ) ⊃ . . . will becomestationary, say at some exponent i0 , and there is an element y ∈ R/p such thatxi0 = xi0 +1 · y. Since R/p is an integral domain, this implies 1 = x · y and, hence,that x is a unit. Therefore R/p is a field and p a maximal ideal in R. Thus, allprime ideals of R are maximal and it follows with the help of 1.3/4 that theJacobson radical of R coincides with the nilradical. Finally, assume that there is an infinite sequence p1 , p2 , p3 , . . . of pairwisedistinct prime ideals in R. Then p1 ⊃ p 1 ∩ p 2 ⊃ p 1 ∩ p 2 ∩ p 3 ⊃ . . .is a descending sequence of ideals in R and, thus, must become stationary, say atsome index i0 . The latter means p1 ∩ . . . ∩ pi0 ⊂ pi0 +1 and we see from 1.3/8 thatone of the prime ideals p1 , . . . , pi0 will be contained in pi0 +1 . However, this isimpossible, since the pi are pairwise distinct maximal ideals in R. Consequently,there exist only finitely many prime ideals in R. In Section 2.4 we will introduce the Krull dimension dim R of a ring R. Itis defined as the supremum of all integers n such that there is a chain of primeideals p0 . . . pn of length n in R. At the moment we are only interested inrings of dimension 0, i.e. rings where every prime ideal is maximal. For example,any non-zero Artinian ring R is of dimension 0 by Proposition 7.Theorem 8. For a non-zero ring R the following conditions are equivalent: (i) R is Artinian. (ii) R is Noetherian and dim R = 0.Proof. Since Artinian rings are of dimension 0, we must show that, for ringsof Krull dimension 0, Artinian is equivalent to Noetherian. To do this we useCorollary 6 and show that in a Noetherian or Artinian ring R of dimension 0the zero ideal is a product of maximal ideals. Assume first that R is a Noetherian ring of dimension 0. Then we see from2.1/12 applied to the ideal a = 0 that there are only finitely many prime idealsp1 , . . . , pn in R and that the intersection of these is the nilradical rad(R). Thelatter is finitely generated and, hence, nilpotent, say (rad(R))r = 0 for someinteger r > 0. Then (p1 · . . . · pn )r ⊂ (p1 ∩ . . . ∩ pn )r = rad(R)r = 0shows that the zero ideal in R is a product of maximal ideals.
  16. 16. 70 2. The Theory of Noetherian Rings Now consider the case, where R is Artinian. Then R contains only finitelymany prime ideals p1 , . . . , pn by Proposition 7 and all of these are maximal inR. In particular, we get p1 ∩ . . . ∩ pn = rad(R) from 1.3/4. As exercised in theNoetherian case, it is enough to show that the nilradical rad(R) is nilpotent. To do this, we proceed indirectly. Writing a = rad(R), we assume thatai = 0 for all i ∈ N. Since R is Artinian, the descending chain a1 ⊃ a2 ⊃ . . .becomes stationary, say at some index i0 . Furthermore, let b ⊂ ai0 be a (non-zero) minimal ideal such that ai0 b = 0. We claim that p = (0 : ai0 b) is a primeideal in R. Indeed, p = R, since ai0 b = 0. Furthermore, let x, y ∈ R such thatxy ∈ p. Assuming x ∈ p, we get ai0 bxy = 0, but ai0 bx = 0. Then bx = b bythe minimality of b and therefore ai0 by = 0 so that y ∈ p. Thus, p is prime andnecessarily belongs to the set of prime ideals p1 , . . . , pn . In particular, we haveai0 ⊂ p. But then ai0 b = ai0 ai0 b ⊂ pai0 b = 0,which contradicts the choice of b. Therefore it follows that a = rad(R) is nilpo-tent and we are done.Exercises 1. Let R be a Noetherian local ring with maximal ideal m. Show that R/q is Artinian for every m-primary ideal q ⊂ R. n 2. Let R1 , . . . , Rn be Artinian rings. Show that the cartesian product i=1 Ri is Artinian again. 3. Let R be an Artinian ring and let p1 , . . . , pr be its (pairwise different) prime ideals. Show: (a) The canonical homomorphism R E r n i=1 R/pi is an isomorphism if n is large enough. (b) The canonical homomorphisms Rpi E R/pn , i = 1, . . . , r, are isomorphisms i if n is large enough. Consequently, the isomorphism of (a) can be viewed as a canonical isomorphism R ∼ E r i=1 Rpi . 4. Let A be an algebra of finite type over a field K, i.e. a quotient of type K X1 , . . . , Xn /a by some ideal a of the polynomial ring K X1 , . . . , Xn . Show that A is Artinian if and only if the vector space dimension dimK A is finite. Hint: Use Exercise 3 in conjunction with the fact to be proved in 3.2/4 that dimK A/m < ∞ for all maximal ideals m ⊂ A. 5. Call an R-module M simple if it admits only 0 and M as submodules. A Jordan–H¨lder sequence for an R-module M consists of a decreasing sequence o of R-modules M = M0 ⊃ M1 ⊃ . . . ⊃ Mn = 0 such that all quotients Mi−1 /Mi , i = 1, . . . , n, are simple. Show that an R-module M admits a Jordan–H¨lder o sequence if and only if M is Artinian and Noetherian.
  17. 17. 2.3 The Artin–Rees Lemma 712.3 The Artin–Rees LemmaWe will prove the Artin–Rees Lemma in order to derive Krull’s Intersection The-orem from it. The latter in turn is a basic ingredient needed for characterizingthe Krull dimension of Noetherian rings in Section 2.4.Lemma 1 (Artin–Rees Lemma). Let R be a Noetherian ring, a ⊂ R an ideal,M a finitely generated R-module, and M ⊂ M a submodule. Then there existsan integer k ∈ N such that (ai M ) ∩ M = ai−k (ak M ∩ M )for all exponents i ≥ k. Postponing the proof for a while, let us give some explanations concerningthis lemma. The descending sequence of ideals a1 ⊃ a2 ⊃ . . . defines a topologyon R, the so-called a-adic topology; see 6.1/3 for the definition of a topology.Indeed, a subset E ⊂ R is called open if for every element x ∈ E there existsan exponent i ∈ N such that x + ai ⊂ E. Thus, the powers ai for i ∈ N arethe basic open neighborhoods of the zero element in R. In a similar way, onedefines the a-adic topology on any R-module M by taking the submodules ai Mfor i ∈ N as basic open neighborhoods of 0 ∈ M . Now if M is a submodule ofM , we may restrict the a-adic topology on M to a topology on M by takingthe intersections ai M ∩ M as basic open neighborhoods of 0 ∈ M . Thus, asubset E ⊂ M is open if and only if for every x ∈ E there exists an exponenti ∈ N such that x + (ai M ∩ M ) ⊂ E. However, on M the a-adic topology exists as well and we may try to com-pare both topologies. Clearly, since ai M ⊂ ai M ∩ M , any subset E ⊂ M thatis open with respect to the restriction of the a-adic topology on M to M will beopen with respect to the a-adic topology on M . Moreover, in the situation ofthe Artin–Rees Lemma, both topologies coincide, as follows from the inclusions (ai M ) ∩ M = ai−k (ak M ∩ M ) ⊂ ai−k Mfor i ≥ k. A topology on a set X is said to satisfy the Hausdorff separation axiomif any different points x, y ∈ X admit disjoint open neighborhoods. Since twocosets with respect to ai in R, or with respect to ai M in M are disjoint as soonas they do not coincide, it is easily seen that the a-adic topology on R (resp.M ) is Hausdorff if and only if we have i∈N ai = 0 (resp. i∈N ai M = 0). Incertain cases, the latter relations can be derived from the Artin–Rees Lemma:Theorem 2 (Krull’s Intersection Theorem). Let R be a Noetherian ring, a ⊂ Ran ideal such that a ⊂ j(R), and M a finitely generated R-module. Then ai M = 0. i∈N
  18. 18. 72 2. The Theory of Noetherian RingsProof. Applying the Artin–Rees Lemma to the submodule M = i∈N ai M ofM , we obtain an index k ∈ N such that M = ai−k M for i ≥ k. Since M asa finitely generated module over a Noetherian ring is Noetherian by 1.5/12, wesee that the submodule M ⊂ M is finitely generated. Therefore Nakayama’sLemma 1.4/10 yields M = 0. Now in order to prepare the proof of the Artin–Rees Lemma, consider aNoetherian ring R, an ideal a ⊂ R, and an R-module M of finite type togetherwith an a-filtration (Mi )i∈N . By the latter we mean a descending sequence ofsubmodules M = M0 ⊃ M1 ⊃ . . . such that ai Mj ⊂ Mi+j for all i, j ∈ N. Thefiltration (Mi )i∈N is called a-stable if there is an index k such that Mi+1 = a · Mifor all i ≥ k or, equivalently, if Mk+i = ai · Mk for all i ∈ N. Furthermore, ilet us consider the direct sum R• = i∈N a as a ring by using component-wise addition and by distributively extending the canonical multiplication mapsai × aj E ai+j for i, j ∈ N that are given on components. In a similar way,we can view the direct sum M• = i∈N Mi as an R• -module. Note that R• isa graded ring in the sense of 9.1/1 and M• a graded R• -module in the sense ofSection 9.2. Observe that the ring R• is Noetherian. Indeed, R is Noetherian and, hence,the ideal a admits a finite system of generators a1 , . . . , an . Thus, the canonicalinjection R E R• extends to a surjection R X1 , . . . , Xn E R• by sendingthe polynomial variable Xν to aν for ν = 1, . . . , n. At this point aν has to beviewed as an element in the first power a1 , which is the component of index 1 inR• . Since the polynomial ring R X1 , . . . , Xn is Noetherian by Hilbert’s BasisTheorem 1.5/14, we see that R• is Noetherian as well.Lemma 3. For a Noetherian ring R, an ideal a ⊂ R, and an R-module Mof finite type, let R• and M• be as before. Then the following conditions areequivalent: (i) The filtration (Mi )i∈N is a-stable. (ii) M• is an R• -module of finite type.Proof. Starting with the implication (i) =⇒ (ii), let k be an integer such thatMk+i = ai · Mk for all i ∈ N. Then M• is generated as an R• -module bythe subgroup i≤k Mi ⊂ M• . Since M is a finitely generated R-module overa Noetherian ring, it is Noetherian by 1.5/12. Hence, all its submodules Miare finitely generated and the same is true for the finite direct sum i≤k Mi .Choosing a finite system of R-generators for the latter, it generates M• as anR• -module. Conversely, assume that M• is a finitely generated R• -module. Then wecan choose a finite system of “homogeneous generators”, namely of generatorsx1 , . . . , xn , where xν ∈ Mσ(ν) for certain integers σ(1), . . . , σ(n). It follows thataMi = Mi+1 for i ≥ max{σ(1), . . . , σ(n)} and, hence, that the filtration (Mi )i∈Nof M is a-stable.
  19. 19. 2.3 The Artin–Rees Lemma 73 Now the proof of the Artin–Rees Lemma is easy to achieve. Consider thefiltration (Mi )i∈N of M given by Mi = ai M ; it is a-stable by its definition.Furthermore, consider the induced filtration (Mi )i∈N on M ; the latter is givenby Mi = Mi ∩ M . Then M• is canonically an R• -submodule of M• . Since M•is of finite type by the Lemma 3 and R• is Noetherian, we see from 1.5/12 thatM• is of finite type as well. Thus, by Lemma 3 again, the filtration (Mi )i∈N isa-stable, and there exists an integer k such that a(ai M ∩ M ) = ai+1 M ∩ Mfor i ≥ k. Iteration yields ai (ak M ∩ M ) = ak+i M ∩ M for i ∈ N and, thus, theassertion of the Artin–Rees Lemma.Exercises 1. Let M be an R-module. Verify for an ideal a ⊂ R that the a-adic topology on M , as defined above, is a topology in the sense of 6.1/3. Show that the a-adic topology on M satisfies the Hausdorff separation axiom if and only if i∈N ai M = 0. 2. Consider the a-adic topology on an R-module M for some ideal a ⊂ R. Show that a submodule N ⊂ M is closed in M if and only if i∈N (ai M + N ) = N . For R Noetherian, M finitely generated and a ⊂ j(R), deduce that every submodule of M is closed. 3. Generalization of Krull’s Intersection Theorem: Consider an ideal a of a Noethe- rian ring R and a finitely generated R-module M . Then i∈N ai M consists of all elements x ∈ M that are annihilated by 1 + a for some element a ∈ a. In partic- ular, show i∈N ai = 0 for any proper ideal a of a Noetherian integral domain. Hint: Use the generalized version of Nakayama’s Lemma in Exercise 1.4/5. 4. For a field K and an integer n > 1 consider the homomorphism of formal power series rings σ : K X E K X given by f (X) E f (X n ). Show that σ is an injective homomorphism between local rings and satisfies σ(X) ⊂ (X) for the maximal ideal (X) ⊂ K X . Construct a ring R as the union of the infinite chain of rings σ σ σ K X ⊂ E K X ⊂ E K X ⊂ E ... and show that R is a local integral domain where Krull’s Intersection Theorem is not valid any more. In particular, R cannot be Noetherian. Hint: For the definition of formal power series rings see Exercise 1.5/8. 5. Given an ideal a ⊂ R, show that gra (R) = i i+1 is canonically a ring; i∈N a /a it is called the graded ring associated to a. Show that gra (R) is Noetherian if R is Noetherian. Furthermore, let M = M0 ⊃ M1 ⊃ . . . be an a-filtration on an R-module M . Show that gr(M ) = i∈N Mi /Mi+1 is canonically a gra (R)-module and that the latter is finitely generated if M is a finite R-module and the filtration on M is a-stable.
  20. 20. 74 2. The Theory of Noetherian Rings2.4 Krull DimensionIn order to define the dimension of a ring R, we use strictly ascending chainsp0 p1 . . . pn of prime ideals in R where the integer n is referred to as thelength of the chain.Remark 1. Let R be a ring and p ⊂ R a prime ideal. Then: (i) The chains of prime ideals in R starting with p correspond bijectively tothe chains of prime ideals in R/p starting with the zero ideal. (ii) The chains of prime ideals in R ending with p correspond bijectively tothe chains of prime ideals in the localization Rp ending with pRp .Proof. Assertion (i) is trivial, whereas (ii) follows from 1.2/5.Definition 2. For a ring R, the supremum of lengths n of chains p0 p1 ... pn ,where the pi are prime ideals in R, is denoted by dim R and called the Krulldimension or simply the dimension of R. For example, fields are of dimension 0, whereas a principal ideal domain isof dimension 1, provided it is not a field. In particular, we have dim Z = 1, aswell as dim K X = 1 for the polynomial ring over a field K. Also we know thatdim K X1 , . . . , Xn ≥ n, since the polynomial ring K X1 , . . . , Xn contains thechain of prime ideals 0 (X1 ) (X1 , X2 ) . . . (X1 , . . . , Xn ). In fact, we willshow dim K X1 , . . . , Xn = n in Corollary 16 below. Likewise, the polynomialring K X1 , X2 , . . . in an infinite sequence of variables is of infinite dimension,whereas the zero ring 0 is a ring having dimension −∞, since by conventionthe supremum over an empty subset of N is −∞. Any non-zero ring containsat least one prime ideal and therefore is of dimension ≥ 0.Definition 3. Let R be a ring. (i) For a prime ideal p ⊂ R its height ht p is given by ht p = dim Rp . (ii) For an ideal a ⊂ R, its height ht a is given by the infimum of all heightsht p, where p varies over all prime ideals in R containing a. (iii) For an ideal a ⊂ R, its coheight coht a is given by coht a = dim R/a. Thus, for a prime ideal p ⊂ R, its height ht p (resp. its coheight coht p)equals the supremum of all lengths of chains of prime ideals in R ending at p(resp. starting at p). In particular, we see that ht p + coht p ≤ dim R. In order to really work with the notions of dimension and height, it isnecessary to characterize these in terms of so-called parameters. For example,we will show that an ideal a of a Noetherian ring satisfies ht a ≤ r if it can begenerated by r elements; see Krull’s Dimension Theorem 6 below. Furthermore,we will prove that any Noetherian local ring R of dimension r admits a system of
  21. 21. 2.4 Krull Dimension 75r elements, generating an ideal whose radical coincides with the maximal idealm ⊂ R; see Proposition 11 below. If m itself can be generated by r elements,we face a special case, namely, where R is a so-called regular local ring. To prepare the discussion of such results, we need a special case of Krull’sIntersection Theorem 2.3/2.Lemma 4. Let R be a Noetherian ring and p ⊂ R a prime ideal. Then, lookingat the canonical map R E Rp from R into its localization by p, we have ker(R E Rp ) = p(i) , i∈Nwhere p(i) = pi Rp ∩ R is the ith symbolic power of p, as considered at the endof Section 2.1.Proof. Since Rp is a local ring with maximal ideal pRp by 1.2/7 and sincewe can conclude from 1.2/5 (ii) that Rp is Noetherian, Krull’s IntersectionTheorem 2.3/2 is applicable and yields i∈N pi Rp = 0. Using the fact thattaking preimages with respect to R E Rp is compatible with intersections,we are done. Now we can prove a first basic result on the height of ideals in Noetherianrings, which is a special version of Krull’s Principal Ideal Theorem to be derivedin Corollary 9 below. For any ideal a ⊂ R we consider the set of minimal primedivisors of a; thereby we mean the set of prime ideals p ⊂ R that are minimalamong those satisfying a ⊂ p. In the setting of 2.1/12, this is the set Ass (a)of isolated prime ideals associated to a. Also note that, for a prime ideal pcontaining a there is always a minimal prime divisor of a contained in p. Thisfollows from 2.1/12 (ii) or without using the theory of primary decompositionsby applying Zorn’s Lemma.Lemma 5. Let R be a Noetherian integral domain and consider a non-zeroelement a ∈ R that is not a unit. Then ht p = 1 for every minimal primedivisor p of (a).Proof. For a minimal prime divisor p of (a), we can pass from R to its local-ization Rp and thereby assume that R is a Noetherian local integral domainwith maximal ideal p. Then we have to show that any prime ideal p0 ⊂ R thatis strictly contained in p satisfies p0 = 0. To do this, look at the descendingsequence of ideals (i) ai = p0 + (a), i ∈ N, (i)where p0 is the ith symbolic power of p0 , and observe that R/(a) is Artinian by2.2/8. Indeed, R/(a) is Noetherian and satisfies dim R/(a) = 0, as p is the onlyprime ideal containing a. Therefore the sequence of the ai becomes stationary,say at some index i0 . Hence,
  22. 22. 76 2. The Theory of Noetherian Rings (i ) (i) p0 0 ⊂ p0 + (a), i ≥ i0 ,and, in fact (i ) (i) (i ) p0 0 = p0 + ap0 0 , i ≥ i0 , (i ) (i )using the equality (p0 0 : a) = p0 0 ; the latter relation follows from 2.1/7, since (i )p0 0 is p0 -primary and since a ∈ p0 . Now Nakayama’s Lemma in the version of (i ) (i)1.4/11 implies p0 0 = p0 for i ≥ i0 . Furthermore, since R is an integral domain,Lemma 4 shows p0 = ker(R E Rp0 ) = 0. (i ) (i) p0 0 = i∈N (i )But then we conclude from pi0 ⊂ p0 0 that p0 = 0 and, thus, that ht p = 1. 0Theorem 6 (Krull’s Dimension Theorem). Let R be a Noetherian ring anda ⊂ R an ideal generated by elements a1 , . . . , ar . Then ht p ≤ r for everyminimal prime divisor p of a.Proof. We conclude by induction on the number r of generators of a. The caser = 0 is trivial. Therefore let r > 0 and consider a minimal prime divisor p of awith a strictly ascending chain of prime ideals p0 . . . pt = p in R. It has tobe shown that t ≤ r. To do this, we can assume that R is local with maximalideal p and, hence, that p is the only prime ideal in R containing a. Furthermore,we may suppose t > 0 and, using the Noetherian hypothesis, that there is noprime ideal strictly between pt−1 and pt . Since a ⊂ pt−1 , there exists a generatorof a, say ar , that is not contained in pt−1 . Then p = pt is the only prime idealcontaining pt−1 + (ar ), and it follows from 1.3/6 (iii) that rad(pt−1 + (ar )) = p.Thus, there are equations of type an = ai + ar yi , i i = 1, . . . , r − 1,for some exponent n > 0 where ai ∈ pt−1 and yi ∈ R. Now consider the ideala = (a1 , . . . , ar−1 ) ⊂ R. Since a ⊂ pt−1 , there is a minimal prime divisor p ofa such that p ⊂ pt−1 . By the above equations we have p = rad(a) ⊂ rad a + (ar ) ⊂ rad p + (ar ) ⊂ p.In particular, p is a minimal prime divisor of p + (ar ) or, in other words, p/p isa minimal prime divisor of ar · R/p in R/p . But then we get ht(p/p ) = 1 fromTheorem 5 and we can conclude p = pt−1 from p ⊂ pt−1 pt = p. Thus, pt−1turns out to be a minimal prime divisor of a . Since the latter ideal is generatedby r − 1 elements, we get t − 1 ≤ r − 1 by induction hypothesis and, thus, aredone. Let us list some immediate consequences from Krull’s Dimension Theorem:Corollary 7. Let a be an ideal of a Noetherian ring R. Then ht a < ∞.
  23. 23. 2.4 Krull Dimension 77Corollary 8. Let R be a Noetherian local ring with maximal ideal m. Thendim R ≤ dimR/m m/m2 < ∞.Proof. First note that dimR/m m/m2 < ∞, since R is Noetherian and, hence, m isfinitely generated. Choosing elements a1 , . . . , ar ∈ m giving rise to an R/m-basisof m/m2 , we conclude from Nakayama’s Lemma in the version of 1.4/12 thatthe ai generate m. But then dim R = ht m ≤ r = dimR/m m/m2 by Krull’sDimension Theorem.Corollary 9 (Krull’s Principal Ideal Theorem). Let R be a Noetherian ring anda an element of R that is neither a zero divisor nor a unit. Then every minimalprime divisor p of (a) satisfies ht p = 1.Proof. We get ht p ≤ 1 from Krull’s Dimension Theorem. Since the minimalprime ideals in R are just the isolated prime ideals associated to the zero ideal0 ⊂ R by 2.1/12, the assertion follows from the characterization of zero divisorsgiven in 2.1/10. Next, to approach the subject of parameters, we prove a certain converse ofKrull’s Dimension Theorem.Lemma 10. Let R be a Noetherian ring and a an ideal in R of height ht a = r.Assume there are elements a1 , . . . , as−1 ∈ a such that ht(a1 , . . . , as−1 ) = s − 1for some s ∈ N where 1 ≤ s ≤ r. Then there exists an element as ∈ a such thatht(a1 , . . . , as ) = s. In particular, there are elements a1 , . . . , ar ∈ a such that ht(a1 , . . . , ar ) = r.Proof. We have only to justify the first statement, as the second one follows fromthe first by induction. Therefore consider elements a1 , . . . , as−1 ∈ a, 1 ≤ s ≤ r,such that ht(a1 , . . . , as−1 ) = s − 1, and let p1 , . . . , pn ⊂ R be the minimalprime divisors of (a1 , . . . , as−1 ). Then, using Krull’s Dimension Theorem, wehave ht pi ≤ s − 1 < r for all i. Since ht a = r, this implies a ⊂ pi for alli and, hence, a ⊂ n pi by 1.3/7. Thus, choosing as ∈ a − n pi , we get i=1 i=1ht(a1 , . . . , as ) ≥ s and, in fact ht(a1 , . . . , as ) = s by Krull’s Dimension Theorem. For the discussion of parameters, which follows below, recall from 2.1/3 thatan ideal a of a local ring R with maximal ideal m is m-primary if and only ifrad(a) = m.Proposition 11. Let R be a Noetherian local ring with maximal ideal m. Thenthere exists an m-primary ideal in R generated by d = dim R elements, but nosuch ideal that is generated by less than d elements.Proof. Combine Lemma 10 with Krull’s Dimension Theorem 6.
  24. 24. 78 2. The Theory of Noetherian RingsDefinition 12. Let R be a Noetherian local ring with maximal ideal m. A setof elements x1 , . . . , xd ∈ m is called a system of parameters of R if d = dim Rand the ideal (x1 , . . . , xd ) ⊂ R is m-primary. In particular, we see from the Proposition 11 that every Noetherian localring admits a system of parameters.Proposition 13. Let R be a Noetherian local ring with maximal ideal m. Then,for given elements x1 , . . . , xr ∈ m, the following conditions are equivalent: (i) The system x1 , . . . , xr can be enlarged to a system of parameters of R. (ii) dim R/(x1 , . . . , xr ) = dim R − r. Furthermore, conditions (i) and (ii) are satisfied if ht(x1 , . . . , xr ) = r.Proof. First observe that condition (i) follows from ht(x1 , . . . , xr ) = r, usingLemma 10. Next consider elements y1 , . . . , ys ∈ m. Writing R = R/(x1 , . . . , xr )and m = m/(x1 , . . . , xr ), the residue classes y 1 , . . . , y s generate an m-primaryideal in R if and only if x1 , . . . , xr , y1 , . . . , ys generate an m-primary ideal in R.In particular, if x1 , . . . , xr , y1 , . . . , ys is a system of parameters of R, as we mayassume in the situation of (i), we have r + s = dim R and can conclude fromProposition 11 that dim R ≤ dim R−r. On the other hand, if the residue classesof y1 , . . . , ys form a system of parameters in R, we have s = dim R and see that(x1 , . . . , xr , y1 , . . . , ys ) is an m-primary ideal in R. Therefore dim R ≤ r + dim Rby Proposition 11. Combining both estimates yields dim R ≤ dim R−r ≤ dim Rand, thus, dim R = dim R − r, as needed in (ii). Conversely, assume (ii) and consider elements y1 , . . . , ys ∈ m, whose residueclasses form a system of parameters in R so that s = dim R. Then the systemx1 , . . . , xr , y1 , . . . , ys generates an m-primary ideal in R. Since r + s = dim R by(ii), the system is, in fact, a system of parameters and (i) follows.Corollary 14. Let R be a Noetherian local ring with maximal ideal m, and leta ∈ m be an element that is not a zero divisor. Then dim R/(a) = dim R − 1.Proof. Use the fact that ht(a) = 1 by Corollary 9. As an application of the theory of parameters, let us discuss the dimensionof polynomial rings.Proposition 15. Consider the polynomial ring R X1 , . . . , Xn in n variablesover a Noetherian ring R. Then dim R X1 , . . . , Xn = dim R + n.Proof. We will show dim R X = dim R + 1 for one variable X, from whichthe general case follows by induction. Let p0 . . . pr be a strictly ascendingchain of prime ideals in R of length r. Then p0 R X ... pr R X pr R X + XR X
  25. 25. 2.4 Krull Dimension 79is a strictly ascending chain of prime ideals of length r + 1. From this we seethat dim R X ≥ dim R + 1. To show that this is actually an equality, considera maximal ideal m ⊂ R X and set p = m ∩ R. Then p is a prime ideal in Rand it is enough to show that ht m ≤ ht p + 1. To do this, we replace R by itslocalization Rp and interpret the polynomial ring Rp X as the localization ofR X by the multiplicative system R − p. Thereby we have reduced our claimto the case where R is a Noetherian local ring with maximal ideal p; use 1.2/7.Then R/p is a field and R X /pR X R/p X a principal ideal domain.Hence, there is a polynomial f ∈ m such that m = pR X + f R X .Now let x1 , . . . , xr be a system of parameters of R where r = dim R. Thenx1 , . . . , xr , f generate an ideal in R X whose radical is m. Therefore Theorem 6shows ht m ≤ r + 1 = ht p + 1, as claimed.Corollary 16. dim K X1 , . . . , Xn = n for the polynomial ring in n variablesover a field K. In particular, we can conclude for finitely generated K-algebras, i.e. quo-tients of polynomial rings of type K X1 , . . . , Xn , that their dimension is finite.Making use of the fact to be proved later in 3.2/4 that any maximal idealm ⊂ K X1 , . . . , Xn leads to a residue field K X1 , . . . , Xn /m that is finiteover K, we can even derive the following more specific version of Corollary 16:Proposition 17. Consider the polynomial ring K X1 , . . . , Xn in n variablesover a field K and a maximal ideal m ⊂ K X1 , . . . , Xn . Then: (i) m is generated by n elements. (ii) ht m = n. (iii) The localization K X1 , . . . , Xn m is a local ring of dimension n, whosemaximal ideal is generated by a system of parameters.Proof. To establish (i) and (ii), we use induction on n, the case n = 0 beingtrivial. So assume n ≥ 1. Let n = m ∩ K X1 , . . . , Xn−1 and consider theinclusions K ⊂ E K X1 , . . . , Xn−1 /n ⊂ E K X1 , . . . , Xn /m.Then K X1 , . . . , Xn /m is a field that is finite over K by 3.2/4 and, hence,finite over K X1 , . . . , Xn−1 /n. In particular, using 3.1/2, the latter is a field aswell. Therefore n is a maximal ideal in K X1 , . . . , Xn−1 , and we may assumeby induction hypothesis that n, as an ideal in K X1 , . . . , Xn−1 , is generatedby n − 1 elements. Now look at the canonical surjection K X1 , . . . , Xn−1 /n Xn E K X1 , . . . , Xn /m
  26. 26. 80 2. The Theory of Noetherian Ringssending Xn onto its residue class in K X1 , . . . , Xn /m. Since on the left-hand side we are dealing with a principal ideal domain, there is a polynomialf ∈ K X1 , . . . , Xn such that m is generated by n and f . In particular, m isgenerated by n elements and assertion (i) is clear. Furthermore, we see thatht m ≤ n by Krull’s Dimension Theorem 6. On the other hand, since K X1 , . . . , Xn /(n) K X1 , . . . , Xn−1 /n Xn ,it is clear that n generates a prime ideal in K X1 , . . . , Xn different from m sothat n m. Using ht n = n − 1 from the induction hypothesis, we get ht m ≥ nand, hence, ht m = n, showing (ii). Finally, (iii) is a consequence of (i) and (ii). To end this section, we briefly want to touch the subject of regular localrings.Proposition and Definition 18. For a Noetherian local ring R with maximalideal m and dimension d the following conditions are equivalent: (i) There exists a system of parameters of length d in R generating themaximal ideal m; in other words, m is generated by d elements. (ii) dimR/m m/m2 = d. R is called regular if it satisfies the equivalent conditions (i) and (ii).Proof. Assume (i) and let x1 , . . . , xd be a system of parameters generatingm. Then, as an R/m-vector space, m/m2 is generated by d elements and,hence, dimR/m m/m2 ≤ d. Since dim R ≤ dimR/m m/m2 by Corollary 8, we getdimR/m m/m2 = d and therefore (ii). Conversely, assume (ii). Then m/m2 is generated by d elements and the sameis true for m by Nakayama’s Lemma in the version of 1.4/12. In particular, sucha system of generators is a system of parameters then. For example, we see from Hilbert’s Basis Theorem 1.5/14 in conjunctionwith Proposition 17 that for a polynomial ring K X1 , . . . , Xn over a field Kits localization K X1 , . . . , Xn m at a maximal ideal m is an example of a regularNoetherian local ring of dimension n. It is known that any localization Rp of aregular Noetherian local ring R by a prime ideal p ⊂ R is regular again; see Serre[24], Prop. IV.23. Also, let us mention the Theorem of Auslander–Buchsbaumstating that any regular Noetherian local ring is factorial; see Serre [24], Cor. 4of Thm. IV.9 for a proof of this fact. To deal with some more elementary properties of regular local rings, it isquite convenient to characterize Krull dimensions of rings not only via lengthsof ascending chains of prime ideals, or systems of parameters, as we have done,but also in terms of the so-called Hilbert polynomial. For example, see Atiyah–Macdonald [2], Chapter 11, for such a treatment. From this it is easily seen thatregular Noetherian local rings are integral domains. Since we need it later on,we will prove this fact by an ad hoc method.
  27. 27. 2.4 Krull Dimension 81Proposition 19. Let R be a regular Noetherian local ring. Then R is an integraldomain.Proof. We argue by induction on d = dim R. Let m be the maximal ideal of R.If d = 0, then m is generated by 0 elements and, hence, m = 0. Therefore R is afield. Now let dim R > 0 and let x1 , . . . , xd be a system of parameters generatingm. Furthermore, let p1 , . . . , pr be the minimal prime ideals in R; their numberis finite by 2.1/12. We claim that we can find an element of type d a = x1 + ci xi ∈ m − m 2 i=2for some coefficients ci ∈ R such that a is not contained in any of the minimalprime ideals p1 , . . . , pr . Clearly, any element a of this type cannot be containedin m2 since x1 , . . . , xd give rise to an R/m-vector space basis of m/m2 . Using a recursive construction for a, assume that a ∈ p1 , . . . , pt for somet < r, but that a ∈ pt+1 . Applying 1.3/8, there exists an element t c∈ pj − pt+1 j=1and we can find an index i0 ∈ {2, . . . , d} such that cxi0 ∈ pt+1 . Otherwise,since c ∈ pt+1 and cxi ∈ pt+1 implies xi ∈ pt+1 , all elements a, x2 , . . . , xd wouldbe contained in pt+1 so that m ⊂ pt+1 . However, this contradicts the fact thatdim R = d > 0. Then we see that a = a + cxi0 is of the desired type: ais not contained in p1 , . . . , pt , since this is true for a and since c ∈ t pj . j=1Furthermore, a ∈ pt+1 , since a ∈ pt+1 and cxi0 ∈ pt+1 . Thus, we have seen that there exists an element a ∈ m − m2 as specifiedabove that is not contained in any of the minimal prime ideals p1 , . . . , pr of R.Now look at the quotient R/(a). Its dimension is d − 1 by Proposition 13, sincea, x2 , . . . , xd form a system of parameters of R. Furthermore, x2 , . . . , xd giverise to a system of parameters of R/(a) generating the maximal ideal of thisring. Therefore R/(a) is regular and, thus, by induction hypothesis, an integraldomain. In particular, we see that the ideal (a) ⊂ R is prime. Now considera minimal prime ideal of R that is contained in (a), say p1 ⊂ (a). Then anyelement y ∈ p1 is of type ab for some b ∈ R and, in fact b ∈ p1 , since a ∈ p1 .Therefore p1 = (a) · p1 and Nakyama’s Lemma 1.4/10 shows p1 = 0. Hence, Ris an integral domain.Exercises 1. Consider the polynomial ring R X in one variable over a not necessarily Noethe- rian ring. Show dim R + 1 ≤ dim R X ≤ 2 · dim R + 1. Hint: Let p1 p2 ⊂ R X be two different prime ideals in R X restricting to the same prime ideal p ⊂ R. Deduce that p1 = pR X .
  28. 28. 82 2. The Theory of Noetherian Rings2. Let K X, Y be the polynomial ring in two variables over a field K. Show dim K X, Y /(f ) = 1 for any non-zero polynomial f ∈ K X, Y that is not constant.3. Let R be a Noetherian local ring with maximal ideal m. Show for a1 , . . . , ar ∈ m that dim R/(a1 , . . . , ar ) ≥ dim R − r. Hint: Assume r = 1. Show for any chain of prime ideals p0 ... pn where a1 ∈ pn that there is a chain of prime ideals p1 . . . pn satisfying a1 ∈ p1 and pn = pn ; use induction on n.4. Consider the formal power series ring R = K X1 , . . . , Xn in finitely many vari- ables over a field K. Show that R is a regular Noetherian local ring of dimension n. Hint: Use Exercise 1.5/8 for the fact that R is Noetherian.5. Let R be a regular Noetherian local ring. Show that R is a field if dim R = 0. Show that R is a discrete valuation ring, i.e. a local principal ideal domain, if dim R = 1.6. Let R be a regular Noetherian local ring of dimension d with maximal ideal m ⊂ R. Show for elements a1 , . . . , ar ∈ m that the quotient R/(a1 , . . . , ar ) is regular of dimension d − r if and only if the residue classes a1 , . . . , ar ∈ m/m2 are linearly independent over the field R/m.7. Let R be a regular Noetherian local ring of dimension d with maximal ideal m. Let m = (a1 , . . . , ad ) and set k = R/m. Show for polynomial variables X1 , . . . , Xd that the canonical k-algebra homomorphism k X1 , . . . , Xd E gr (R) = mi /mi+1 , Xj E aj , m i∈N where aj is the residue class of aj in m/m2 , is an isomorphism of k-algebras. Hints: See Exercise 2.3/5 for the fact that the graded ring grm (R) associated to m is Noetherian if R is Noetherian. Proceed by induction, similarly as in the proof of Proposition 19, and look at the maximal ideal i>0 mi /mi+1 ⊂ grm (R). Show that this maximal ideal cannot be a minimal prime ideal in grm (R) if d = dim R > 0.
  29. 29. http://www.springer.com/978-1-4471-4828-9

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